A 25.0mL sample of 2.0mol/L sodium oxalate is reacted with 30.0mL of 1.50 mol/L copper (II) nitrate solution a student recovers 5.82 grams o

Question

A 25.0mL sample of 2.0mol/L sodium oxalate is reacted with 30.0mL of 1.50 mol/L copper (II) nitrate solution a student recovers 5.82 grams of precipitate when they perform the reaction.

Write a balenced chemical reaction.

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Dulcie 4 years 2021-07-20T19:42:54+00:00 1 Answers 19 views 0

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  1. Vodka
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    2021-07-20T19:44:21+00:00
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    Answer:

    Na2C2O4(aq) + Cu(NO3)2(aq) → 2NaNO3(aq) + CuC2O4(s)

    Explanation:

    Sodium oxalate, Na2C2O4 reacts with copper (II) nitrate, Cu(NO3)2 producing an insoluble salt. As the reaction is of 2 salts you can know there is an ion exchange reaction where sodium nitrate is produced ans copper (II) oxalate (This salt is the precipitate). The reaction is:

    Na2C2O4(aq) + Cu(NO3)2(aq) → 2NaNO3(aq) + CuC2O4(s)

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )